WORK POWER ENERGY

1. Work, Energy, and Power
We utilized Newton's laws to analyze the motion of objects. Force and mass information were
used to determine the acceleration of an object. Acceleration information was subsequently
used to determine information about the velocity or displacement of an object after a given
period of time.
Motion will be approached from the perspective of work and energy. The effect that work has
upon the energy of an object (or system of objects) will be investigated; the resulting velocity
and/or height of the object can then be predicted from energy information.

2. WORK
◦ When a force acts upon an object to cause a displacement of the object, it is said that work was done upon the object.
◦ There are three key ingredients to work - force, displacement, and cause.
◦ In order for a force to qualify as having done work on an object, there must be a displacement and the force
must cause the displacement.
◦ There are several good examples of work that can be observed in everyday life –
◦ a horse pulling a plow through the field,
◦ a father pushing a grocery cart down the aisle of a grocery store,
◦ a freshman lifting a backpack full of books upon her shoulder,
◦ a weightlifter lifting a barbell above his head,
◦ an Olympian launching the shot-put, etc.
In each case described here there is a force exerted upon an object to cause that object to be displaced.

3. ◦ In Physics, work is done when a force acts on an object as the moves from one place to another.
◦ Because of the work, energy is transferred from the object that causes the force to the object on which the force acts, or vice
versa.
Work Equation
W = F • d • cos 𝜽
Where:
F is the force,
d is the displacement, and
The angle 𝜽 (theta) is defined as the angle between the force and the displacement vector.
Perhaps the most difficult aspect of the above equation is the angle "theta." The angle is not just any 'ole angle, but rather a very
specific angle. The angle measure is defined as the angle between the force and the displacement.

4. Units of Work
◦ The units of work are the product of the units of force (newton or pounds) and the units of displacement (meters of feet). Thus
the work and energy units are newton-meters (N and m) or pound-feet (lb. and ft.)
◦ Work and energy’s units has been given a special name in the SI metric system, it is called joule (J) in honor of the English
Physicist James Joule (1818-1889).
Work (and also energy), the standard metric unit is the Joule (abbreviated J). One Joule is equivalent to one Newton of force
causing a displacement of one meter. In other words,
The Joule is the unit of work.
1 Joule = 1 Newton * 1 meter
1 J = 1 N * m
1 J = 1 Nm

5. Example
1. A carpenter pulls a cart load of lumber 10 m across a floor. The tension in the rope is 200 N and is directed 37° above the horizontal.
Calculate the work done by the carpenter in pulling the lumber.
Solution:
The tension force T and displacement r make an angle of 37 with respect to each other. Hence, the work done by the rope tension force in
moving the land of lumber is
W = F d cos θ
W = (200 N)(10 m) cos 37°
W = (200 N)(10 m) (0.80)
W = 1,600 J

6. 2. A woman jumps from a high wall and lands in soft sand. The sand exerts and average upward force of 50,000N while stopping
her. The woman stops after sinking 5 cm into the sand. How much work is done by the force of the sand on her?
The force of the sand on the woman points upward, whereas the woman’s displacement while she is stopping is 5cm downward, hence there is a 180
angle between the direction of the force and the direction of the displacement.
5 cm x
1 m
100 cm
= 0.05 m
W = F Δr cosθ
W = (50,000 N)(0.05 m) cos 180°
W = (50,000 N)(0.05 m) (-1)
W = -2,500 Nm
W = -2,500 J
Note: If the negative work is done on an object, the object’s energy decreases.

7. ENERGY
Whenever work is done on an object, it gains energy. If 100J of work is done, the object gains 100J of energy. The energy may be
stored by the object in a variety of forms.
Kinetic Energy
- The energy an object has because it is moving
Gravitational Potential Energy
- The energy an object has because of its vertical separation from the earth.
Elastic Potential Energy
- The energy stored in a stretched or compressed elastic material such as a spring.

8. Kinetic Energy (KE)
Kinetic energy is the energy of motion. An object that has motion - whether it is vertical or horizontal motion - has kinetic energy.
◦ KE = 0.5 • m • v2
where m = mass of object
v = speed of object
This equation reveals that the kinetic energy of an object is directly proportional to the square of its speed. That means that for a
twofold increase in speed, the kinetic energy will increase by a factor of four.
Like work and potential energy, the standard metric unit of measurement for kinetic energy is the Joule. As might be implied by
the above equation, 1 Joule is equivalent to 1 kg*(m/s)^2.
1 Joule = 1 kg • m2/s2

9. Example:
1. Determine the kinetic energy of a 625-kg roller coaster car that is
moving with a speed of 18.3 m/s.
Solution:
KE = (0.5)mv2
KE = (0.5) (625 kg) (18.3 m/s2)
KE = 1.05 x 105 Joules

10. 2. If the roller coaster car in the above problem were moving with twice the speed, then
what would be its new kinetic energy?
Solution:
If the speed is doubles, then KE is quadrupled. Thus, KE = 4 (1.04653 x 105 J) = 4.19 x 105
J.
or
KE = 0.5 mv2
KE = (0.5)(625 kg)(36.6 m/s2)
KE = 4.19x105 J

11. 3. Missy Diwater, the former platform diver for the Ringling Brother's Circus, had a kinetic energy of 12 000 J just prior to hitting
the bucket of water. If Missy's mass is 40 kg, then what is her speed?
Solution:
KE = 0.5 mv2
12,000 J = (0.5) (40 kg) v2
300 J = (0.5) v2
600 J = v2
V = 24.5 m/s

12. The kinetic energy KE of a mass m moving with a speed v is KE =
1
2
mv².
The change in kinetic energy when the speed of the mass changes from an initial valueV0 to a
final is:
◦ ΔKE =KE-KE0=
1
2
mv²-
1
2
mv0² or
◦ ΔKE=
1
2
mvf²-
1
2
mvi²

13. Example
1. A 0.10 kg stone is thrown from the edge on an ocean cliff with an initial speed of 20m/s.
When it strikes the water below, it is traveling at 45 m/s. What is the change in kinetic energy
of the stone?
Given:
◦ m = 0.10 kg
◦ vi = 20 m/s
◦ vf = 45 m/s

14. Solution:
ΔKE=
1
2
mvf²-
1
2
mvi²
ΔKE=
1
2
(0.10 kg)(45 m/s)² -
1
2
(0.10 kg)(20 m/s)²
ΔKE=
1
2
(0.10 kg)(2025 m²/s²) -
1
2
(0.10 kg)(400 m²/s²)
ΔKE= 101. 25 kg m²/s² - 20 kg m²/s²
ΔKE= 81.25 kg (m/s²)m
ΔKE= 81.25 Nm
ΔKE= 81.25 J

15. Potential Energy
◦ An object can store energy as the result of its position. For example, the heavy ball of a demolition
machine is storing energy when it is held at an elevated position. This stored energy of position is
referred to as potential energy.
◦ Similarly, a drawn bow is able to store energy as the result of its position. When assuming its usual
position (i.e., when not drawn), there is no energy stored in the bow. Yet when its position is altered from
its usual equilibrium position, the bow is able to store energy by virtue of its position.
◦ Potential energy is the stored energy of position possessed by an object.

16. Gravitational Potential Energy
◦ Gravitational potential energy is the energy stored in an object as the result of its vertical position or
height. The energy is stored as the result of the gravitational attraction of the Earth for the object.
◦ The gravitational potential energy of the massive ball of a demolition machine is dependent on two
variables - the mass of the ball and the height to which it is raised.
◦ There is a direct relation between gravitational potential energy and the mass of an object. More massive
objects have greater gravitational potential energy.
◦ There is also a direct relation between gravitational potential energy and the height of an object. The
higher that an object is elevated, the greater the gravitational potential energy.

17. Gravitational Potential Energy cont.
◦ These relationships are expressed by the following equation:
PEgrav = mass • g • height
PEgrav = mgh
Where:
m represents the mass of the object, h represents the height of the object and g represents the
gravitational field strength (9.8 N/kg on Earth) - sometimes referred to as the acceleration of gravity.

18. Gravitational Potential Energy cont.
An object has more gravitational potential energy when higher above the earth than when closer
to the earth. In the vertical separation of a mass (m) from the earth is changed from an initial
position (𝑦0) to a final position (y), then there is a changed in gravitational potential energy.
Gravity is a form of energy an object has because of its vertical separation from the
earth.
ΔPEg = PEg- PEg0
= mgy - mgy0 = wy - wy0
or ΔPEg = mg (y - y0)

19. Example:
1. A 400kg roller coaster car starts 48m above the ground and is 3m above the ground at the end of
the ride. Calculate the change in gravitational potential energy of the car.
m y0 = 48 m and yf = 3 m. Hence, the change in gravitational potential energy is:
ΔPEg = mg (y - 𝑦0)
ΔPEg = (400 kg)( 9.8 m/s²) [(3 m) – (48 m)]
ΔPEg = -180, 000 J
Conclusion: At the end of the ride the roller coaster is closer to the earth and so it has less gravitational potential energy.

20. Elastic Potential Energy (𝐏𝐄𝐬)
◦ Elastic potential energy is the energy stored in elastic materials as the result of their stretching or
compressing. Elastic potential energy can be stored in rubber bands, bungee chords, trampolines,
springs, an arrow drawn into a bow, etc. The amount of elastic potential energy stored in such a device is
related to the amount of stretch of the device - the more stretch, the more stored energy.
◦ Springs are a special instance of a device that can store elastic potential energy due to either compression
or stretching. A force is required to compress a spring; the more compression there is, the more force
that is required to compress it further. For certain springs, the amount of force is directly proportional to
the amount of stretch or compression (x); the constant of proportionality is known as the spring constant
(k).
Fspring = k • x

21. Such springs are said to follow Hooke's Law. If a spring is not stretched or compressed, then there
is no elastic potential energy stored in it. The spring is said to be at its equilibrium position. The equilibrium
position is the position that the spring naturally assumes when there is no force applied to it. In terms of
potential energy, the equilibrium position could be called the zero-potential energy position. There is a
special equation for springs that relates the amount of elastic potential energy to the amount of stretch (or
compression) and the spring constant. The equation is
PEspring = 0.5 • k • x2
or Pes = ½ k x2
where: k = spring constant
x = amount of compression (relative to equilibrium position)

22. To summarize, potential energy is the energy that is stored in an object due to its
position relative to some zero position.
An object possesses gravitational potential energy if it is positioned at a height
above (or below) the zero height.
An object possesses elastic potential energy if it is at a position on an elastic
medium other than the equilibrium position.

23. 1. A cart is loaded with a brick and pulled at constant speed along an inclined plane to the
height of a seat-top. If the mass of the loaded cart is 3.0 kg and the height of the seat top is
0.45 meters, then what is the potential energy of the loaded cart at the height of the seat-
top?
Solution:
PE = mgh
PE = (3 kg) (9.8 m/s2) (0.45 m)
PE = 13.2 J

24. 2. If a force of 14.7 N is used to drag the loaded cart (from previous question) along the incline
for a distance of 0.90 meters, then how much work is done on the loaded cart?
Solution:
W = F d cos theta
W = (14.7 N) (0.90 m) (cos 0 degrees)
W = 13.2 J
Note: The angle between F and d is 0 degrees because the F and d are in the same direction.

25. Other Example:
1. The spring in BB gun has a force constant of 1.8 x 104 N/m. When loaded, the spring is compressed a
distance of 1.2 cm. Calculate the change in elastic potential energy of the spring when it is released and
returns to its equilibrium position.
Solution:
ΔPEs= PEs − PE𝑠0
= ½ kx² - ½ kxo
ΔPEs = ½ kx² - ½ kxo²
ΔPEs = ½ (1.8 x 104 N/m) (0 m)² - ½ (1.8 x 104 N/m) (0.012 m)²
ΔPEs = 0 - 1.3 Nm
ΔPEs = -1.3 J
Conclusion:
The negative sign implies that the spring lost elastic energy as it went from its initial compressed position to its final relaxed
position at x = 0.

26. POWER
◦ The quantity work has to do with a force causing a displacement. Work has nothing to do with
the amount of time that this force acts to cause the displacement. Sometimes, the work is
done very quickly and other times the work is done rather slowly.
◦ Power is the rate at which work is done. It is the work/time ratio. Mathematically, it is
computed using the following equation.
Power = Work / time
or P = W / t
The standard metric unit of power is the Watt.
1 watt = 1 Joule/ second

27. Example:
1. A tired squirrel (mass of approximately 1 kg) does push-ups by applying a force to elevate its center-of-mass by 5 cm in order
to do a mere 0.50 Joule of work. If the tired squirrel does all this work in 2 seconds, then determine its power.
The tired squirrel does 0.50 J of work in 2.0 seconds. The power rating of this squirrel is found by
P = W / t
P = (0.50 J) /(2.0 s)
P = 0.25 watts

28. Another Formula for Power
◦ The expression for power is work/time. And since the expression for work is force*displacement, the expression for
power can be rewritten as (force*displacement)/time. Since the expression for velocity is displacement/time, the
expression for power can be rewritten once more as force*velocity.

29. Example:
1. Suppose that Ben Pumpiniron elevates his 80-kg body up the 2.0-meter stairwell in 1.8 seconds. If this were the case,
then we could calculate Ben's power rating. It can be assumed that Ben must apply an 800-Newton downward force
upon the stairs to elevate his body. By so doing, the stairs would push upward on Ben's body with just enough force to lift
his body up the stairs. It can also be assumed that the angle between the force of the stairs on Ben and Ben's
displacement is 0 degrees. With these two approximations, Ben's power rating could be determined as shown below.
Solution:
Power = Work/ time
W = F d cos θ

30. 2. A crane lifts a 300 kg load at constant speed in vertical distance of 30 m in 10 sec. Calculate the rate in
which the crane doing on the land.
Solution:
F = ma = (300 kg) (9.8 m/s²) = 2,940 N
W = Fd = (2,940 N) (30 m) = 88,200 J
P =
𝒘
𝒕
=
88,200 J
10 sec
= 8,820 watts
◦